Effect of pointcharges on electric dipole in an equilateral triangle

[ppy]

2 particles of charge q are placed at 2 vertices of an equilateral triangle of side a. An electric dipole is placed at the third vertex with its dipole moment orientated parallel to the opposite side of the triangle.

a) Determine the magnitude of the torque on the dipole due to the electric field from the 2 point charges?

b) Using F=-∇U where U=-p.E is the potential energy of a dipole moment p in an electric field E determine the magnitude and the direction of the translational force on the dipole due to the electric field from the 2 point charges.

c) Determine the magnitude of the electric field from the dipole at the position of one of the charges


This is what I have however I am very confused.
a) $\tau$=pxE=pEsinθ

For 1 point charge E=q/4$\pi$ε$_{0}$a$^{2}$

$\tau$=2pqsin(45)/4$\pi$ε$_{0}$=$\sqrt{2}$pq/4$\pi$ε$_{0}$

I thought the 2 is needed because you have the 2 point charges of charge q however I don't know can you just add them as it is a magnitude

c) I know that the electric field for a dipole is ($\frac{2pcosθ}{4\piε_{0}}$r$^{3}$, $\frac{psinθ}{4\piε_{0}}$r$^{3}$, 0) in spherical polars.

 

[Adithyan]

First, determine the 'resultant' field at the vertex due to the two charges (vector addition of both the fields- the direction of the resultant field will come out to be upwards, i.e, perpendicular to the base and try to find the magnitude yourself). Now use the formula for torque.

For the last question, resolve the field due to dipole as radial and equatorial - as shown in the diagram. Use the formula for the components.

 

[ppy]

First, determine the 'resultant' field at the vertex due to the two charges (vector addition of both the fields- the direction of the resultant field will come out to be upwards, i.e, perpendicular to the base and try to find the magnitude yourself). Now use the formula for torque.

For the last question, resolve the field due to dipole as radial and equatorial - as shown in the diagram. Use the formula for the components.

so is the resultant electric field $\frac{\sqrt{2}q}{4\piε_{0}a^{2}}$ in the perpendicular direction to the base and therefore the magnitude of the torque is pEsinθ=$\frac{pq}{4\piε_{0}a^{2}}$ ?

What about for b) ?

 

[Adithyan]

so is the resultant electric field $\frac{\sqrt{2}q}{4\piε_{0}a^{2}}$ in the perpendicular direction to the base and therefore the magnitude of the torque is pEsinθ=$\frac{pq}{4\piε_{0}a^{2}}$ ?

No, the resultant is $\frac{q}{4\piε_{0}a^{2}}$ and the magnitude of torque is $\frac{pq}{4\piε_{0}a^{2}}$.

What about for b) ?

I can't really understand the significance of the question. The question asks for the direction of force using potential energy whereas the translational force on a dipole is zero (equal magnitude and opposite charged particles).

(By the way, F=-du/dr)

 

[ppy]

No, the resultant is $\frac{q}{4\piε_{0}a^{2}}$ and the magnitude of torque is $\frac{pq}{4\piε_{0}a^{2}}$.



I can't really understand the significance of the question. The question asks for the direction of force using potential energy whereas the translational force on a dipole is zero (equal magnitude and opposite charged particles).

(By the way, F=-du/dr)

Thankyou for the reply

I thought for a) the electric fields in the x direction will cancel but the electric field for the y direction will be $\frac{qsin(45)}{4\piε_{0}a^{2}}$=$\frac{\sqrt{2}q}{4\piε_{0}a^{2}}$ for one point charge and therefore twice this for 2 point charges as the y components are in the same direction. What am I doing wrong? Then when we calculate the torque=pEsinθ=pEsin(45)=$\frac{pq}{4\piε_{0}a^{2}}$

Why is my resultant electric field incorrect?

 

[Adithyan]

Θ=90 sinΘ=1